Question

Two rods of same material and volume having circular cross-section are subjected to tension T. Within the elastic limit, same force is applied to both the rods. Diameter of the first rod is half of the second rod, then the extensions of first rod to second rod will be in the ratio

विकल्प

• 4 : 1

• 32 : 1

• 16 : 1

• 2 : 1

Answer 1

2 : 1

Explanation:

Rods have the same Young's modulus because they are formed of the same material.

Y₁ = Y₂

${\displaystyle \frac{\frac{{\text{F}}_1}{{\text{A}}_1}}{\frac{\Delta {\text{L}}_1}{{\text{L}}_1}}=\frac{\frac{{\text{F}}_2}{{\text{A}}_2}}{\frac{\Delta {\text{L}}_2}{{\text{L}}_2}}}$

${\displaystyle \frac{{\text{F}}_1{\text{L}}_1}{\Delta {\text{L}}_1\times 2\pi \left(\frac{{\text{d}}_1}{2}\right){\text{L}}_1}=\frac{{\text{F}}_2{\text{L}}_2}{\Delta {\text{L}}_2\times 2\pi \left(\frac{{\text{d}}_2}{2}\right){\text{L}}_2}}$

${\displaystyle \frac{{\text{F}}_1}{\Delta {\text{L}}_1\times {\text{d}}_1}=\frac{{\text{F}}_2}{\Delta {\text{L}}_2\times {\text{d}}_2}}$

${\displaystyle \frac{\Delta {\text{L}}_1}{\Delta {\text{L}}_2}=\frac{{\text{F}}_1}{{\text{d}}_1}\times \frac{{\text{d}}_2}{{\text{F}}_2}}$

Here, F₁ = F₂ = F and d₁ = ${\displaystyle \frac{{\text{d}}_2}{2}}$

∴ ${\displaystyle \frac{\Delta {\text{L}}_1}{\Delta {\text{L}}_2}=\frac{2}{1}}$ or 2 : 1